0
$\begingroup$

DISCLAIMER: Newbie/Wannabe Physicist.

What's to stop us from making a "Gravity Mill" which is a bar/rod attached with weights on both ends and pivoted at the centre with a fulcrum? Similar to Newton's cradle which keeps swinging, this Gravity Mill would keep rotating (under ideal physical circumstances) due to gravitational force, thereby allowing us to attach a dynamo and extract power from this force. Please correct me if I am wrong anywhere.

$\endgroup$
3
$\begingroup$

You're right that the rod would keep swinging under ideal physical circumstances, but this wouldn't be due to gravity. If you imagine the rod at the horizontal point of its rotation, there is a gravitational force on one side pulling on one weight, trying to speed up its rotation, and a gravitational force on the other side pulling on the other weight, trying to slow down its rotation. In the language of physics, we say that the net torque is zero (or that the net moment is zero). What this means is that gravity does not have any effect on the rod, and the rod would spin just as if gravity weren't there.

The fact that the rod keeps spinning indefinitely is a statement of the conservation of angular momentum, which states that when there is no torque acting on a rigid object, its speed of rotation is constant. Its the rotational analogue of the conservation of linear momentum, which states that when there is no net force acting on an object, its velocity is constant.

Now why can't we extract power from this rotating rod? To answer this, imagine that the physical circumstances were not ideal. For instance, imagine that there was some friction at the pivot. This friction acts as a torque on the spinning rod, such that its speed of rotation decreases, rather than stays constant. The effect is to transfer energy out of the rod, which slows down, and into the bearings and the pivot, which warm up (that is, gain thermal energy). The total energy is conserved.

Attaching a dynamo to the pivot, in circumstances without friction, has the same effect. Ultimately you want your dynamo to generate electrical energy in the wires connected to it. To do that, you need to draw energy from some other source - in this case, the spinning wheel. This is the law of conservation of energy, and it implies that our rod will slow down until it eventually stops.

$\endgroup$
  • $\begingroup$ Thank you very much for the explanations and potential enlightenment on these concepts. As for keeping the wheel/device in perpetual motion, I may have over-simplified the apparatus. Perfect examples would be DaVinci's overbalanced wheel and other similar devices. Please check out the following videos: [youtube.com/watch?v=287qd4uI7-E] [youtube.com/watch?v=WQEbXD4Kqwc] [youtube.com/watch?v=xDF0cugCoMM] $\endgroup$ – prgSRR Jan 23 '15 at 8:57
  • $\begingroup$ Given that we do have perpetual motion devices like these, are you still indicating that practically power cannot be generated from these (Due to law of conservation of energy)? $\endgroup$ – prgSRR Jan 23 '15 at 9:03
  • 1
    $\begingroup$ @prgSRR we don't have perpetual motion devices. Possibly you might like to take some of those videos to skeptics.stackexchange... $\endgroup$ – Flyto Jan 24 '15 at 10:33
  • $\begingroup$ Sorry, I haven't found time to watch all those videos, but I can confirm that indeed, no practical power can be generated from devices like these. The law of conservation of energy is an incredibly powerful law: it means that even without knowing any of the specifics of these machines, we can categorically say that they can't produce energy. Some of these machines work by optical illusion, some generate useful power for a rather long time but eventually slow down and stop.. In principle, some machines could even continue indefinitely, but no energy could be extracted from them without slowing. $\endgroup$ – gj255 Jan 26 '15 at 21:31
  • $\begingroup$ @SimonW: Thanks for the insights. I'll do further research on these supposedly perpetual motion machines. $\endgroup$ – prgSRR Jan 27 '15 at 10:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.